A 4.40–kg block is set into motion up an inclined plane with an initial speed of

Question

A 4.40–kg block is set into motion up an inclined plane with an initial speed of vi = 8.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of = 30.0° to the horizontal.

(a) For this motion, determine the change in the block's kinetic energy.
J

(b) For this motion, determine the change in potential energy of the block–Earth system.
J

(c) Determine the friction force exerted on the block (assumed to be constant).
N

(d) What is the coefficient of kinetic friction?

Explanation / Answer

Here ,

a)

change in kientic energy = final kinetic enrgy - initial KE

change in kientic energy = -0.5 * 4.4 * 8.4^2

change in kientic energy = -155.2 J

the change in kientic energy is -155.2 J

b)

change in potential energy = mgh

change in potential energy = 4.4 * 9.8 * 3 * sin(30)

change in potential energy = 64.68 J

the change in potential energy is 64.68 J

c)

let the frictional force is f

for the work done by friction

f * d = change in KE + change in PE

-f * 3 = -155.2 + 64.68

f = 30.2 N

the frictional force f is 30.2 N

d)

let the coefficient of friction is u

u * mg * cos(theta) = f

30.2 = u * 4.4 * 9.8 * cos(30)

u = 0.81

the coeffcient of friction is 0.81

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